An elementary lemma of commutative algebra - MathOverflow

An elementary lemma of commutative algebra

2010-10-01T04:53:52Z

Let $R$ be a commutative ring, $M$, $N$ $R$-modules, and $f: M\rightarrow N$ a homomorphism. It is known that $f$ is injective (surjective) if and only if $f_m$ is injective (surjective) for all maximal ideal $m$. But I don't know whether $f$ is split if and only if $f_m$ is split? Maybe it is true for finitely generated modules over noetherian rings, I suspect.

Answer by Emerton for An elementary lemma of commutative algebra

2010-10-01T05:16:26Z

Let $0 \to A \to B \to C \to 0$ be a short exact sequence of $R$-modules, with $R$ Noetherian and $C$ of finite type, giving rise to a class $c \in Ext^1_R(C,A)$. Tensoring with $R_{\mathfrak m}$ for a maximal ideal $\mathfrak m$ gives the short exact sequence $0 \to A_{\mathfrak m} \to B_{\mathfrak m} \to C_{\mathfrak m} \to 0$, which corresponds to the image of $c$ in the $Ext^1_{R_{\mathfrak m}}(C_{\mathfrak m},A_{\mathfrak m}) = R_{\mathfrak m}\otimes_R Ext^1_R(C,A)$. (This isomorphism holds because $R_{\mathfrak m}$ is flat over $R$; to prove it one computes the Ext via a resolution of $C$ by finite rank free $R$-modules, which exists since $R$ is Noetherian and $C$ is of finite type.)

Now suppose that these localizations are all split. Then we have an element $c$ in the $R$-module $Ext^1_R(C,A)$ whose image in the localization at each maximal ideal $\mathfrak m$ vanishes. It follows that this element itself vanishes. (Its annihilator is not contained in any maximal ideal.)

So the answer seems to be "yes": for Noetherian $R$, if a short exact sequence whose third term is finite type splits locally at every $\mathfrak m$, it splits.

Note: I think one can also see this by comparing sheaf Exts with actual Exts for the associated quasi-coherent sheaves on Spec $R$. The point is that there is a spectral sequence involving the cohohomology of the sheaf Exts converging to the actual Exts (in some generality), but on the affine Spec $R$ the higher cohomology of the sheaf Exts vanishes, so that a class in any $Ext^i$ is a global section of the sheaf Ext, and hence is determined by what happens locally. (Hopefully this is not all nonsense.)

Answer by Lennart Galinat for An elementary lemma of commutative algebra

2010-10-01T13:45:12Z

If you want to avoid the use of Ext-groups, you could prove it like this (which is basically the same proof):

Let 0→A→B→C→0 be a short exact sequence of R-modules with C finitely presented and assume it splits after localisation at every maximal ideal.

Use the natural isomorphism Hom$_{R_m}(C_m,A_m)$=R$_m$⊗Hom$_R$(C,A) (which uses the flatness of localisation and the finitely presentedness of C) and the assumption to see that the map Hom$_R$(C,B)→Hom$_R$(C,C) is surjective since all of its localisations are.

Since the giving a splitting is equivalent to this map being surjective we are done.

P.S.: Of course one could prove "if and only if" in the statement like this.

Counterexample in the general case: Take R=$\prod_{\mathbb N} \mathbb F_2$, I=$\sum_{\mathbb N} \mathbb F_2$ and C the cokernel of the inclusion.

Now I claim three things: 1.) R has dimension zero

Proof: Every element of R is an idempotent and every prime ideal in R has to contain exactly one of e or 1-e for every idempotent e in R. If we had a chain of prime ideals, the larger one would nescessarily have to contain e and 1-e for one such idempotent and so couldn't exist.

2.) The localisation of R at every maximal ideal is a field.

Proof: It is a zero-dimensional local ring by 1.) and reduced since R doesn't contain nilpotent elements. Thus this localisation is a field.

3.) I is not a direct summand of R.

Proof: Direct Summands correspond to idempotents in R. Since every element in R is idempotent we need to analyze all principal ideals. If an element has only finitely many non-zero entries the ideal created by it has only finitely many elements, thus can't be I. If on the other hand it contains infinitely many non-zero entries, the ideal created by it has uncountably many elements. Since I contains countably many elements, it can't be a direct summand of R.

This establishes the counterexample, since the short exact sequence 0→I→R→C→0 splits in every localisation at a maximal ideal.